[seqfan] Re: Help with a calculation for A001097 (twin primes)

[Bob Selcoe]

Hi Alan,

Thanks for the reply.  Another respondent (off-list) ran a program 
verifying, among other things, the limit at -inf. (should have seen that); 
I've been reformulating the problem to eliminate that issue. But for now 
this question has been answered.

Cheers,
Bob




--------------------------------------------------
From: "Allan Wechsler" <acwacw at gmail.com>
Sent: Wednesday, November 19, 2014 10:19 AM
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Subject: [seqfan] Re: Help with a calculation for A001097 (twin primes)

> Bob, what do you want to have happen when sum(1/a(n)) goes above 1? Your
> multiplicand will go negative and the sum will start to drop. It's not
> clear to me that it has a well-defined limit.
>
>
> On Wed, Nov 19, 2014 at 10:26 AM, Bob Selcoe <rselcoe at entouchonline.net>
> wrote:
>
>> Hello Seqfans,
>>
>> Is there a way to calculate the following limit as n approaches infinity?
>> This will be helpful in a comment I intend to add to A001097 (twin 
>> primes)
>>
>> Not sure about the most efficient or best way to notate, so here's what
>> I'm asking:
>>
>> For a(n)>1, let a(n) = A007310 (numbers congruent to 1 or 5 mod 6): 5, 7,
>> 11, 13, 17, 19, 23, 25....
>>
>> So a(1) = 5, a(2) = 7, a(3) = 11 etc.
>>
>> Calculate 1/a(1)  +  1/a(2)*(1 - 1/a(1))  +  1/a(3)*(1 - 1/a(1) - 1/a(2))
>> +  1/a(4)*(1 - 1/a(1) - 1/a(2) - 1/a(3))...
>>
>> So 1/5 + 4/35 + 23/385 + 218/5005 +...
>>
>> If this is not easily solved, could someone please run a program to get 
>> an
>> approximation?
>>
>> Thanks in advance!
>>
>>  Cheers,
>> Bob Selcoe
>>
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>>
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>>
>
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